HMMT 十一月 2018 · 冲刺赛 · 第 11 题

HMMT November 2018 — Guts Round — Problem 11

[ 8 ] Let 4 ABC be an acute triangle, with M being the midpoint of BC , such that AM = BC . Let D and E be the intersection of the internal angle bisectors of ∠ AM B and ∠ AM C with AB and AC , respectively. Find the ratio of the area of 4 DM E to the area of 4 ABC .

解析

[ 8 ] Let 4 ABC be an acute triangle, with M being the midpoint of BC , such that AM = BC . Let D and E be the intersection of the internal angle bisectors of \ AM B and \ AM C with AB and AC , respectively. Find the ratio of the area of 4 DM E to the area of 4 ABC . Proposed by: Handong Wang 2 Answer: 9 Let [ XY Z ] denote the area of 4 XY Z . AD AE 2 Solution 1: Let AM = , let DE = d , and let the midpoint of DE be F . Since = = AB AC 3 by the angle bisector theorem, F lies on AM and 4 ADE is similar to 4 ABC . Note that \ DM E is formed by angle bisectors of \ AM B and \ AM C , which add up to 180 . Thus \ DM E is right, so d both 4 DM F and 4 EM F are isosceles. This implies that F M = . Applying the similarity between 2 d AD 2 4 ADE and 4 ABC , we get = = . Since AF = 2 F M , [ ADE ] = 2[ DM E ]. Finally, since AB 3 2 [ ADE ] [ DM E ] 4 1 4 = , = · = . [ ABC ] 9 [ ABC ] 2 9 9 [ DM E ] [ ADE ] [ DBM ] [ EM C ] AD Solution 2: We compute the ratio by finding 1 . Since = [ ABC ] [ ABC ] [ ABC ] [ ABC ] DB [ ADE ] AE AM 2 2 4 = = 2 by the angle bisector theorem, we see that = = . Also, since 1 EC [ ABC ] 3 3 9 BC 2 [ DBM ] [ EM C ] [ DM E ] [ ADE ] 1 1 1 1 1 1 1 BM = CM = BC , = = , and = = . Thus, = 1 2 [ ABC ] 3 2 6 [ ABC ] 2 3 6 [ ABC ] [ ABC ] 2 [ DBM ] [ EM C ] 4 1 1 = 1 = . [ ABC ] [ ABC ] 9 6 6 9